What is the axis y^2-10y-12x+37=0 ?

The axis y^2-10y-12x+37=0 is y=5

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axis y^2-10y-12x+37=0

axis

y^{2} - 10 y - 12 x + 37 = 0

y = 5

Solution steps

y^{2} - 10 y - 12 x + 37 = 0

Rewrite

y^{2} - 10 y - 12 x + 37 = 0

in the standard form

4 \cdot 3 (x - 1) = (y - 5)^{2}

Therefore parabola properties are:

(h, k) = (1, 5), p = 3

Parabola is of the form

4 p (x - h) = (y - k)^{2}

and is symmetric around the

x

-axis

Axis of symmetry is a line parallel to the

x

-axis which intersects the vertex:

y = 5

y = - 12 x^{2} - 29 x + 230

y = - \frac{63}{16} + \frac{1}{16} x^{2} - \frac{1}{8} x

y = x^{2} + 8 x - 5

y = 3 x^{2} - 5 x 7

y = 7 x^{2} - 10 x + 5